TITLE: Integral inequality similar to Hardy's inequality
QUESTION [7 upvotes]: I am not very sure if that's research level, I hope you don't find it too elementary for this place.
I am trying to solve the following puzzle:
We are given a real function $f$, where $f(x) \geq 0$ and $F := \int_0^x f(t) dt$ and some real $p>1$.
Does $\int_0^\infty f(x)^p e^{-x}dx < \infty$ imply $\int_0^\infty F(x)^pe^{-x}dx < \infty$ ? 
My current approach was modifying proofs of Hardy's theorem. This approach has not been very successful. I keep running into hard to compute integrals that I would need to bound. I hope somebody could maybe provide me with some direction in which I should look. If somebody was interested I could provide details of calculations so far.

REPLY [2 votes]: The question has been already solved by Paata Ivanisvili. I just want to share my insight.
I think I have found the solution. I will use generalized Hardy's inequality, which unfortunately is a bit like killing a fly with a cannon.
As a source I will cite http://www.encyclopediaofmath.org/index.php/Hardy_inequality where you can find references to the original source. 
Lemma - generalized Hardy's (cited from Encyclopedia of Math).
$$
\int_0^\infty \left| \phi(x) \int_0^x f(t) dt \right|^p dx \leq C \int_0^\infty \left| \psi(x)  f(x) \right|^p dx
$$
$\iff$
$$
\sup_{x > 0} \left[ \int_x^\infty  | \phi(t) |^p dt \right]^{\frac{1}{p}} \left[ \int_0^x | \psi(t) |^{-q} dt \right]^{\frac{1}{q}} < + \infty,
$$ where $\frac{1}{p} + \frac{1}{q} = 1 $. 
After using this inequality, the task comes down to verification of the assumption, which is pretty easy. Let me verify.
Given a function $f \geq 0$ and $F(x) = \int_0^{x} f(t)
dt$ I claim  $\int_0^\infty f(x)^p e^{-x}dx < \infty$ implies
$\int_0^\infty F(x)^pe^{-x}dx < \infty$. 
Proof: Assume $\psi(t) = \phi(t) = e^{- \frac{t}{p} }$. When
$\frac{1}{p} + \frac{1}{q} = 1 $, then $\frac{1}{q} = \frac{p-1}{p}$ and $-q = \frac{p}{1-p}$. We investigate
$$
\sup_{x > 0} \left[ \int_x^\infty  e^{-t} dt \right]^{\frac{1}{p}} \left[
\int_0^x e^{-t \frac{1}{1-p}} dt \right]^{\frac{p-1}{p}} = \sup_{x > 0} e^{- \frac{x}{p}} \left( e^{\frac{x}{p-1}} - 1 \right)^{\frac{p-1}{p}}
=
$$
$$
\sup_{x > 0} \left( e^{- \frac{x}{p-1}} e^{\frac{x}{p-1}} - e^{- \frac{x}{p-1}} \right)^{\frac{p-1}{p}}
=
\sup_{x > 0}
    \left(
        1 -
    e^{- \frac{x}{p-1}}
    \right)^{\frac{p-1}{p}},
$$
which is bounded as $p>1$. Now we apply Hardy and our theorem is verified.